# Thread: linear law question

1. ## linear law question

It is assumed that $\displaystyle x$ and $\displaystyle y$ obey a law of the form $\displaystyle y = \ln(ax^2 + b)$. Pairs of values of $\displaystyle x$ and $\displaystyle y$ are recorded in an experiment and tabulated as follows:

By means of a straight line graph, verify that the law is valid.
Use your graph to estimate the values of $\displaystyle a$ and $\displaystyle b$
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I am not very sure about changing the equation to linear form. Do I change $\displaystyle y = \ln(ax^2 + b)$ into $\displaystyle \frac{e^y}{x} = ax + \frac{b}{x}$?
Please help, any tips would be greatly appreciated, thank you.

2. "By means of a straight line graph, verify that the law is valid."

I'm afraid that, strictly speaking, this is impossible as pictures can only indicate, not prove.

3. I think, in the context of the problem, "indicate" is all that's required. I doubt a rigorous proof is required.

Reply to the OP: If you were to plot the equation on some sort of log plot, you might get what you're after. Are you sure the x term is squared?

4. Further note: after plotting $\displaystyle e^{y}$ versus $\displaystyle x$ in Excel, I can see that there is indeed a quadratic relationship. You can fit a quadratic curve with an $\displaystyle R^{2}$ value of 1.

I don't know what the problem means by "straight line graph".

5. Thank you. For this question, I am supposed to plot and draw a graph after converting the equation from a quadratic one to a linear one, like from y = x^2 + 5x to y/x = x + 5 and then plot y/x against x. And yes, I the x term is squared.
Sorry for the lack of latex, it is 1.45am here at night and I haven't slept since 5am yesterday.

6. Hmm. I would question the wisdom of going "from y = x^2 + 5x to y/x = x + 5 and then plot y/x against x." Yes, you can do it. But why? The equation still isn't linear. Why vainly try to force it?

The way you were thinking of doing it is as good as any, I suppose.

7. Well I am not exactly sure, my textbook says it is to solve problems involving two unknowns in a non-linear equation. This method is called 'linear law' but I don't think it is common outside of my country. Thank you anyway.

8. Ok, then. Have a good one!